course PHY 121
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11:59:00 `q001. We obtain an estimate of the acceleration of gravity by determining the slope of an acceleration vs. ramp slope graph for an object gliding down an incline. Sample data for an object gliding down a 50-cm incline indicate that the object glides down the incline in 5 seconds when the raised end of the incline is .5 cm higher than the lower end; the time required from rest is 3 seconds when the raised end is 1 cm higher than the lower end; and the time from rest is 2 seconds when the raised end is 1.5 cm higher than the lower end. What is the acceleration for each trial?
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RESPONSE --> IN order to calculate the acceleration, we must know the change in velocity. v= 50cm/5s = 10cm/s^2 a = 10cm/s^2 / 5s = 2 cm/s^2 v = 50cm/3s = 16.67 cm/s a = 16.67 cm/s / 3s = 5.6 cm/s^2 v = 50cm/2s = 25cm/s a = 25cm/s / 2s = 12.5 cm/s^2
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12:28:58 We can find the accelerations either using equations or direct reasoning. To directly reason the acceleration for the five-second case, we note that the average velocity in this case must be 50 cm/(5 seconds) = 10 cm/s. Since the initial velocity was 0, assuming uniform acceleration we see that the final velocity must be 20 cm/second, since 0 cm/s and 20 cm/s average out to 10 cm/s. This implies a velocity change of 20 cm/second a time interval of 5 seconds, or a uniform acceleration of 20 cm/s / (5 s) = 4 cm/s^2. The acceleration in the 3-second case could also be directly reasoned, but instead we will note that in this case we have the initial velocity v0 = 0, the time interval `dt = 3 sec, and the displacement `ds = 50 cm. We can therefore find the acceleration from the equation `ds = v0 `dt + .5 a `dt^2. Noting first that since v0 = 0 the term v0 `dt must also be 0,we see that in this case the equation reduces to `ds = .5 a `dt^2. We easily solve for the acceleration, obtaining a = 2 `ds / `dt^2. In this case we have a = 2 * (50 cm) / (3 sec)^2 = 11 cm/s^2 (rounded to nearest cm/s^2). For the 2-second case we can use the same formula, obtaining a = 2 * (50 cm) / (2 sec)^2 = 25 cm/s^2.
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RESPONSE --> ok before I read any further I used the avg. veocity instead of the change in velcity. the final velocity would be double the avg. velocity there for ein the first scenario: vAve = 10cm/s vf = 10cm/s *2 = 20cm/s `dv = vf - v0 `dv = 20cm/s - 0cm/s `dv = 20cm/s a = `dv/`dt a= 20cm/s / 5s a = 4cm/s^2 the 2nd scenario vAve = 16.67cm/s vf = 16.67*2 = 33.34 cm/s `dv = vf -v0 = `dv = 33.34 cm/s - 0cm/s `dv = 33.34 cm/s a = `dv / `dt a = 33.34cm/s / 3s a = 11 cm/s^2 vAve = 25cm/s vf = 25 cm * 2 = 50 cm/s `dv = vf - v0 = `dv = 50 cm/s - 0 = 50 cm/s a = `dv / `dt a = 50 cm/s / 2s a = 25 cm/s^2
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12:30:14 We can find the accelerations either using equations or direct reasoning. To directly reason the acceleration for the five-second case, we note that the average velocity in this case must be 50 cm/(5 seconds) = 10 cm/s. Since the initial velocity was 0, assuming uniform acceleration we see that the final velocity must be 20 cm/second, since 0 cm/s and 20 cm/s average out to 10 cm/s. This implies a velocity change of 20 cm/second a time interval of 5 seconds, or a uniform acceleration of 20 cm/s / (5 s) = 4 cm/s^2. The acceleration in the 3-second case could also be directly reasoned, but instead we will note that in this case we have the initial velocity v0 = 0, the time interval `dt = 3 sec, and the displacement `ds = 50 cm. We can therefore find the acceleration from the equation `ds = v0 `dt + .5 a `dt^2. Noting first that since v0 = 0 the term v0 `dt must also be 0,we see that in this case the equation reduces to `ds = .5 a `dt^2. We easily solve for the acceleration, obtaining a = 2 `ds / `dt^2. In this case we have a = 2 * (50 cm) / (3 sec)^2 = 11 cm/s^2 (rounded to nearest cm/s^2). For the 2-second case we can use the same formula, obtaining a = 2 * (50 cm) / (2 sec)^2 = 25 cm/s^2.
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RESPONSE --> Well, after looking at the formulas you used it made sense, but I got the same answers. Is that okay. It made more sense to me to do it the way I did.
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12:32:57 `q002. What are the ramp slopes associated with these accelerations?
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RESPONSE --> the acceleration of 4cm/s^2 is associated with the ramp slope of .5 cm. the acceleration of 11cm/s^2 is associated with the ramp slope of 1.0 cm. the acceleration of 25cm/s^2 is associated with the ramp slope of 1.5 cm.
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12:45:40 For the 5-second trial, where acceleration was 4 cm/s^2, the 'rise' of the ramp was .5 cm and the 'run' was nearly equal to the 50-cm length of the ramp so the slope was very close to .5 cm / (50 cm) = .01. For the 3-second trial, where acceleration was 11 cm/s^2, the 'rise' of the ramp was 1 cm and the 'run' was very close to the 50-cm length, so the slope was very close to 1 cm / (50 cm) = .02. For the 2-second trial, where the acceleration was 25 cm/s^2, the slope is similarly found to be very close to 1.5 cm / (50 cm) = .03.
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RESPONSE --> I misunderstood the question I thought you were just asking for the height of the ramp for each. The rise would be the height of the ramp and the run would the the length of the ramp. .5cm / 50cm = .01 for the 4cm/s^2 acceleration 1 cm / 50 cm = .02 for the 11cm/s^2 accel. 1.5 cm / 50 cm = .03 for the 25 cm/s^2
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12:56:56 `q003. Sketch a reasonably accurate graph of acceleration vs. ramp slope and give a good description and interpretation of the graph. Be sure to include in your description how the graph points seem to lie with respect to the straight line that comes as close as possible, on the average, to the three points.
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RESPONSE --> I put the acceleration on the y-axis and the ramp slope on the x-axis. All three of the graph point almost touch the straight line. The graph appears to be increasing at a constant rate.
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12:57:32 The graph will have acceleration in cm/s^2 on the vertical axis (the traditional y-axis) and ramp slope on the horizontal axis (the traditional x-axis). The graph points will be (.01, 4 cm/s^2), (.02, 11.1 cm/s^2), (.03, 25 cm/s^2). The second point lies somewhat lower than a line connecting the first and third points, so the best possible line will probably be lower than the first and third points but higher than the second. The graph indicates that acceleration increases with increasing slope, which should be no surprise. It is not clear from the graph whether a straight line is in fact the most appropriate model for the data. If timing wasn't particularly accurate, these lines could easily be interpreted as being scattered from the actual linear behavior due to experimental errors. Or the graph could indicate acceleration vs. ramp slope behavior that is increasing at an increasing rate.
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RESPONSE -->
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12:57:36 The graph will have acceleration in cm/s^2 on the vertical axis (the traditional y-axis) and ramp slope on the horizontal axis (the traditional x-axis). The graph points will be (.01, 4 cm/s^2), (.02, 11.1 cm/s^2), (.03, 25 cm/s^2). The second point lies somewhat lower than a line connecting the first and third points, so the best possible line will probably be lower than the first and third points but higher than the second. The graph indicates that acceleration increases with increasing slope, which should be no surprise. It is not clear from the graph whether a straight line is in fact the most appropriate model for the data. If timing wasn't particularly accurate, these lines could easily be interpreted as being scattered from the actual linear behavior due to experimental errors. Or the graph could indicate acceleration vs. ramp slope behavior that is increasing at an increasing rate.
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RESPONSE --> ok
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13:11:16 `q004. Carefully done experiments show that for small slopes (up to a slope of about .1) the graph appears to be linear or very nearly so. This agrees with theoretical predictions of what should happen. Sketch a vertical line at x = .05. Then extend the straight line you sketched previously until it intersects the y axis and until it reaches past the vertical line at x = .05. What are the coordinates of the points where this line intersects the y-axis, and where it intersects the x =.05 line? What are the rise and the run between these points, and what therefore is the slope of your straight line?
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RESPONSE --> (.05, 47) I do not see where the line intersects the y axis.
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13:17:15 A pretty good straight line goes through the points (0, -6 cm/s^2) and (.05, 42 cm/s^2). Your y coordinates might differ by a few cm/s^2 either way. For the coordinates given here, the rise is from -6 cm/s^2 to 42 cm/s^2, a rise of 48 cm/s^2. The run is from 0 to .05, a run of .05. The slope of the straight line is approximately 48 cm/s^2 / .05 = 960 cm/s^2. Note that this is pretty close to the accepted value, 980 cm/second^2, of gravity. Carefully done, this experiment will give us a very good estimate of the acceleration of gravity.
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RESPONSE --> weel I thought that I should have extended the line downward. When I do that, it interscts at the point (0,7) The rise would be from -7cm/s^2 to 47 cm/s^2 a rise of 54 cm/s^2. The run = .05 - 0 = .05cm The slope of the line is approx 54 cm/s^2 / .05 = 1080 cm/s^2.
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13:32:19 `q005. The most accurate way to measure the acceleration of gravity is to use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) for the period of a pendulum. Use your washer pendulum and time 100 complete back-and-forth cycles of a pendulum of length 30 cm. Be sure to count carefully and don't let the pendulum swing out to a position more than 10 degrees from vertical. How long did it take, and how long did each cycle therefore last?
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RESPONSE --> 132 seconds 132s / 100 = 1.32 seconds for each cycle
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13:33:33 100 cycles of a pendulum of this length should require approximately 108 seconds. This would be 108 seconds per 100 cycles, or 108 sec / (100 cycles) = 1.08 sec / cycle. If you didn't count very carefully or didn't time very accurately, you might differ significantly from this result; differences of up to a couple of cycles are to be expected.
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RESPONSE --> I think I let the pendulum go to far.
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14:00:54 `q006. You now have values for the period T and the length L, so you can use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) to find the acceleration g of gravity. Solve the equation for g and then use your values for T and L to determine the acceleration of gravity.
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RESPONSE --> T = 2 `pi / sqrt(g) * sqrt (L) T * sqrt (g) = 2 `pi / sqrt (g) * sqrt (g) * sqrt (L) T * sqrt (g) = 2 `pi * sqrt (L) T/ T * sqrt (g) = 2 `pi * sqrt (L) / T sqrt (g) = 2 `pi * sqrt (L) / T g = 4 `pi^2 * 30cm / 132s g = 4 * 9.86 * 30 cm / 132s g = 39.44 * 30 cm / 132 s g = 1183.2 cm / 132 s g = 8.96 cm/s
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14:07:11 Solving T = 2 `pi / `sqrt(g) * `sqrt(L) for g, we can first multiply both sides by `sqrt(g) to obtain T * `sqrt(g) = 2 `pi `sqrt(L). Then dividing both sides by T we obtain `sqrt(g) = 2 `pi `sqrt(L) / T. Squaring both sides we finally obtain {}g = 4 `pi^2 L / T^2. Plugging in the values given here, L = 30 cm and T = 1.08 sec, we obtain g = 4 `pi^2 * 30 cm / (1.08 sec)^2 = 1030 cm/s^2. You should check these calculations for accuracy, since they were mentally approximated.
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RESPONSE --> I forgot to square the T. See Correction: T = 2 `pi / sqrt(g) * sqrt (L) T * sqrt (g) = 2 `pi / sqrt (g) * sqrt (g) * sqrt (L) T * sqrt (g) = 2 `pi * sqrt (L) T/ T * sqrt (g) = 2 `pi * sqrt (L) / T sqrt (g) = 2 `pi * sqrt (L) / T^2 g = 4 `pi^2 * 30cm / (1.32s)^2 g = 4 * 9.86 * 30 cm / 1.74 s^2 g = 39.44 * 30 cm / 1.74 s^2 g = 1183.2 cm / 1.74 s^2 g = 680 cm/s
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